$\dfrac{ 3a + b }{ -9 } = \dfrac{ 4a + 7c }{ 3 }$ Solve for $a$.
Multiply both sides by the left denominator. $\dfrac{ 3a + b }{ -{9} } = \dfrac{ 4a + 7c }{ 3 }$ $-{9} \cdot \dfrac{ 3a + b }{ -{9} } = -{9} \cdot \dfrac{ 4a + 7c }{ 3 }$ $3a + b = -{9} \cdot \dfrac { 4a + 7c }{ 3 }$ Reduce the right side. $3a + b = -{9} \cdot \dfrac{ 4a + 7c }{ {3} }$ $3a + b = -{3} \cdot \left( 4a + 7c \right)$ Distribute the right side $3a + b = -{3} \cdot \left( {4a} + {7c} \right)$ $3a + b = -{12}a - {21}c$ Combine $a$ terms on the left. ${3a} + b = -{12a} - 21c$ ${15a} + b = -21c$ Move the $b$ term to the right. $15a + {b} = -21c$ $15a = -21c - {b}$ Isolate $a$ by dividing both sides by its coefficient. ${15}a = -21c - b$ $a = \dfrac{ -21c - b }{ {15} }$